GATE CS 2019 Q.63 (CS Q.53): Pointers, Arrays and Strings

Consider the following C program:

#include <stdio.h>
int main()
{
    int a[] = {2, 4, 6, 8, 10};
    int i, sum = 0, *b = a + 4;
    for (i = 0; i < 5; i++)
        sum = sum + (*b - i) - *(b - i);
    printf("%d\n", sum);
    return 0;
}

The output of the above C program is ________.

Your answer

Official answer and solution

Official answer: 10
From: Official GATE 2019 answer key

Solution

Evaluate the pointer arithmetic and array accesses inside the loop for each iteration $i = 0, 1, 2, 3, 4$ .

The array $a$ is initialized as:

a = \{2, 4, 6, 8, 10\}

with indices $a[0] = 2, a[1] = 4, a[2] = 6, a[3] = 8, a[4] = 10$ .

The pointer $b$ is initialized to $a + 4$ , which points to $a[4]$ . Thus, $*b = 10$ throughout the execution of the loop.

In the loop for (i = 0; i < 5; i++), the expression added to sum in each iteration is:

(\text{*b} - i) - \text{*(b - i)}

Here:

*b - i evaluates to $10 - i$ .
*(b - i) points to $a[4 - i]$ .

Evaluating each iteration:

$i = 0$ : $(10 - 0) - a[4] = 10 - 10 = 0$
$i = 1$ : $(10 - 1) - a[3] = 9 - 8 = 1$
$i = 2$ : $(10 - 2) - a[2] = 8 - 6 = 2$
$i = 3$ : $(10 - 3) - a[1] = 7 - 4 = 3$
$i = 4$ : $(10 - 4) - a[0] = 6 - 2 = 4$

Adding these values to sum (initially 0):

\text{sum} = 0 + 1 + 2 + 3 + 4 = 10

The program prints 10.

Final answer

Common trap

Confusing *b - i with *(b - i) or reversing the index order.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2019 Computer Science and Information Technology (CS)
Question: 63 of 65, printed as CS Q.53
Section: Computer Science and Information Technology

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